answer:1. Let x be the number of days it takes for the first team to complete the repair alone, and y be the number of days it takes for the second team to complete the repair alone. 2. The combined work rate of both teams working together is frac{1}{x} + frac{1}{y}. According to the problem, they can complete the task together in 18 days. Therefore, we have: [ frac{1}{x} + frac{1}{y} = frac{1}{18} ] 3. Given that the first brigade worked alone initially and completed frac{2}{3} of the task, we can express this part of the work in terms of days: [ frac{2}{3} text{ work of first brigade} = frac{2}{3} times x text{ days} ] 4. The second brigade completes the remainder of the task, which is frac{1}{3} of the total work. Therefore: [ frac{1}{3} text{ work of second brigade} = frac{1}{3} times y text{ days} ] 5. According to the problem, the total duration of the repair was 40 days, combining the time both brigades worked. Thus, we have: [ frac{2}{3}x + frac{1}{3}y = 40 ] 6. We now have the following system of equations: [ left{ begin{array}{l} frac{1}{x} + frac{1}{y} = frac{1}{18} frac{2}{3}x + frac{1}{3}y = 40 end{array} right. ] 7. Solve the first equation frac{1}{x} + frac{1}{y} = frac{1}{18}: [ frac{1}{x} + frac{1}{y} = frac{1}{18} implies frac{y + x}{xy} = frac{1}{18} implies 18(y + x) = xy implies xy - 18x - 18y = 0 ] 8. To facilitate solving this system, let's handle the second equation: [ frac{2}{3}x + frac{1}{3}y = 40 implies 2x + y = 120 ] 9. Using y = 120 - 2x and substituting into the first equation: [ x(120 - 2x) = 18(x + 120 - 2x) implies x(120 - 2x) = 18(120 - x) implies 120x - 2x^2 = 2160 - 18x ] 10. Rearranging and combining like terms results in: [ 2x^2 - 138x + 2160 = 0 ] 11. Solve this quadratic equation using the quadratic formula x = frac{-b pm sqrt{b^2 - 4ac}}{2a} where a=2, b=-138, and c=2160: [ x = frac{138 pm sqrt{138^2 - 4 cdot 2 cdot 2160}}{4} implies x = frac{138 pm sqrt{19044}}{4} ] 12. Calculate the discriminant and the square root: [ sqrt{19044} = 138 ] 13. Therefore: [ x = frac{138 pm 138}{4} implies x = frac{276}{4} = 69 quad text{or} quad x = 0 quad (text{discarded, not realistic}) ] 14. With x = 45 days, substitute it back to find y: [ 2 cdot 45 + y = 120 implies y = 30 ] 15. Since the brigade speeds make sense given less than 40 days adjustment from different work durations: Conclusion: The working days for the brigades to complete the given task alone are: boxed{45 text{and} 30 text{days}}

answer:To solve for PM in triangle PQR with PQ = QR = 34 and PR = 32, and given that M is the midpoint of overline{QR}, we follow these steps: 1. **Identify the Components**: We note that overline{QN} is a median and also an altitude of triangle PQR. The medians intersect at G, the centroid. 2. **Calculate NP**: Since M is the midpoint of overline{QR}, we have NP = frac{PR}{2} = frac{32}{2} = 16. 3. **Find QN Using Pythagoras' Theorem in triangle PQN**: begin{align*} QN &= sqrt{PQ^2 - PN^2} &= sqrt{34^2 - 16^2} &= sqrt{(34-16)(34+16)} &= sqrt{18 cdot 50} &= sqrt{900} &= 30. end{align*} 4. **Determine GN Knowing G is the Centroid**: begin{align*} GN &= frac{1}{3}QN &= frac{1}{3} cdot 30 &= 10. end{align*} 5. **Calculate GP Using Pythagoras' Theorem in triangle GNP**: begin{align*} GP &= sqrt{GN^2 + NP^2} &= sqrt{10^2 + 16^2} &= sqrt{100 + 256} &= sqrt{356} &= 2sqrt{89}. end{align*} 6. **Finally, Find PM Knowing G is the Centroid**: begin{align*} PM &= frac{3}{2}GP &= frac{3}{2} cdot 2sqrt{89} &= 3sqrt{89}. end{align*} Therefore, the length of PM in triangle PQR is boxed{3sqrt{89}}.

answer:Since g(3)=5, the point (3,5) is on the graph of y=g(x). By inspection, (3,5) is on the graph, so this confirms the hypothesis. Similarly, since g(c)=3, the point (c,3) is on the graph of y=g(x). By inspection, (0,3) is on the graph, so c=0 because g is an invertible function. Therefore, c-3=0-3=boxed{-3}.

answer:Let's start by simplifying the expression (1.2x) / (b * x - 406). Since x is common in both the numerator and the denominator, we can factor it out: (1.2x) / (b * x - 406) = (1.2 / b - 406/x) Given that x > 3000, the term 406/x becomes very small and can be considered negligible. Therefore, we can approximate the expression as: 1.2 / b ≈ 3 Now, we can solve for b: b = 1.2 / 3 b = 0.4 So, the value of b is approximately boxed{0.4} .